Ceva’s theorem

Let $ABC$ be a given triangle and $P$ any point of the plane. If $X$ is the intersection point of $AP$ with $BC$, $Y$ the intersection point of $BP$ with $CA$ and $Z$ is the intersection point of $CP$ with $AB$, then

$$\frac{AZ}{ZB}\cdot \frac{BX}{XC}\cdot \frac{CY}{YA}=1.$$

Conversely, if $X,Y,Z$ are points on $BC,CA,AB$ respectively, and if

$$\frac{AZ}{ZB}\cdot \frac{BX}{XC}\cdot \frac{CY}{YA}=1$$

then $AX,BY,CZ$ are concurrent.

Remarks: All the segments are directed segments (that is $AB=-BA$), and so theorem is valid even if the points $X,Y,Z$ are in the prolongations (even at the infinity) and $P$ is any point on the plane (or at the infinity).

Title	Ceva’s theorem
Canonical name	CevasTheorem
Date of creation	2013-03-22 11:57:15
Last modified on	2013-03-22 11:57:15
Owner	drini (3)
Last modified by	drini (3)
Numerical id	16
Author	drini (3)
Entry type	Theorem
Classification	msc 51A05
Related topic	Triangle
Related topic	Median
Related topic	Centroid
Related topic	Orthocenter
Related topic	OrthicTriangle
Related topic	Cevian
Related topic	Incenter
Related topic	GergonnePoint
Related topic	MenelausTheorem
Related topic	ProofOfVanAubelTheorem
Related topic	VanAubelTheorem
Related topic	BisectorsTheorem
Related topic	DirectedSegment